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User blog:P進大好きbot/New Side Nesting Notation
This is an English translation of my Japanese blog post submitted to a Japanese googological event. This is a notation based on the side nesting method in Japanese googology. It might include many typos and errors, e.g. undefined behaviours or obvious infinite loops. I appreciate feedbacks. = Summary = Faery「This time, I'm gonna create an ordinal notation associated to Veblen hierarchy for y'all!」 Unfortunately, this is a notation not associated to Veblen hierarchy. She tried to create an ordinal notation associated to Veblen hierarchy, but made mistakes in the comarison algorithm and the standardness algorithm. Does it well-founded? I do not know. = Notation = I define a recursive set \(T\) of formal strings consisting of \((\), \()\), and \(+\) in the following recursive way: # \(() \in T\). # For any \((a,b,c) \in T^3\), \((abc) \in T\). # For any \((a,b) \in (T \setminus \{()\})^2\), \(a+b \in T\). I put \(0 = ()\). I define a recursive subset \(PT \subset T\) in the following way: # \(0 \notin PT\). # For any \((a,b,c) \in T^3\), \((abc) \in PT\). # For any \((a,b) \in (T \setminus \{0\})^2\), \(a+b \notin PT\). For example, \((000) \in PT\) while \((000) + (000) \in T \setminus PT\). Faery「Variadic Veblen hierarchy smells too complicated, so it's pretty good to start from \(3\)-ary one!」 = Ordering = I simultaneously define recursive \(2\)-ary relations \(s < t\) and \(s \leq t\) on \(T\) in the following recursive way: # \(s \leq t\) is equivalent to \(s < t\) or \(s = t\). # If \(s = 0\), then \(s < t\) is equivalent to \(t \neq 0\). # If \(s \neq 0\) and \(t = 0\), then \(s < t\) does not hold. # Suppose that there exist an \((a,b,c) \in T^3\) and a \((d,e,f) \in T^3\) such that \(s = (abc)\) and \(t = (def)\): ## If \(a = d\), then \(s < t\) is equivalent to that either one of the following holds: ### \(b < e\) and \(c \leq t\). ### \(b = e\) and \(c < f\). ### \(e < b\) and \(s < f\). ## If \(a \neq d\), then \(s < t\) is equivalent to that either one of the following holds: ### \(a < d\) and \(c \leq t\). ### \(d < a\) and \(s < f\). # If there exist an \((a,b,c) \in T^3\) and a \((d,e) \in PT \times (T \setminus \{0\})\) such that \(s = (abc)\) and \(t = d+e\), then \(s < t\) is equivalent to \(s \leq d\). # If there exist an \((a,b) \in PT \times (T \setminus \{0\})\) and a \((d,e,f) \in T^3\) such that \(s = a+b\) and \(t = (def)\), then \(s < t\) is equivalent to the negation of \(t < s\). # If there exist an \((a,b) \in PT \times (T \setminus \{0\})\) and \((d,e) \in PT \times (T \setminus \{0\})\) such that \(s = a+b\) and \(t = d+e\), then \(s < t\) is equivalent to that either one of the following holds: ## \(a < d\). ## \(a = d\) and \(b < e\). Then \(\leq\) forms a total ordering, but does not form a well-ordering. Faery「Somehow weird... Did I make mistakes...? But I've no idea...」 = Standard Form = I put \(1 = (000)\). I define a recursive map \begin{eqnarray*} T \times T & \to & T \\ (a,t) & \mapsto & (a00) \times (t) \end{eqnarray*} in the following recursive way: # If \(t = 0\), then \((a00) \times (t) = 0\). # Suppose that there exists a \((d,e,f) \in T^3\) such that \(t = (def)\). ## Suppose that \(t < (00(00(a00)+1))\). ### If \(t = 1\), then \((a00) \times (t) = (a00)\). ### If \(d = 0\), \(e = 0\), and \(t \neq 1\), then \((a00) \times (t) = (00(a00)+f)\). ### If \(d = 0\) and \(e \neq 0\), then \((a00) \times (t) = (00(a00)+t)\). ### If \(d \neq 0\), then \((a00) \times (t) = (00(a00)+t)\). ## If \((00(00(a00)+1)) \leq t\), then \((a00) \times (t) = t\). # If there exists a \((d,e) \in (T \setminus \{0\}) \times PT\) such that \(t = d+e\), then \((a00) \times (t) = (a00) \times (d) + (a00) \times (e)\). For example, \((100) \times (1+1) = (100) + (100)\). For each \(x \in T\), I define a recursive subset \(OT_x \subset T\) in the following recursive way: # \(0 \in OT_x\). # For any \((a,b,c) \in T^3\), \((abc) \in OT_x\) is equivalent to that all of the following hold: ## \((a,b,c) \in OT_x^3\). ## If \(a = 0\), then \((100) \times (b) \leq x\) and \(b \in OT_{(100) \times (b)}\). ## If \(a \neq 0\), then \((a+100) \times (b) \leq x\) and \(b \in OT_{(a+100) \times (b)}\). ## \(c < (abc)\). # For any \((s,t) \in (T \setminus \{0\}) \times PT\), \(s+t \in OT_x\) is equivalent to that all of the following hold: ## \((s,t) \in OT_x^2\). ## If \(s \in PT\), then \(t \leq s\). ## If there exists an \((a,b) \in T \times PT\) such that \(s = a+b\), then \(t \leq b\). I put \(OT = \{x \in T \mid x \in OT_x\}\). I call an expression in \(OT\) a standard form expression. For example, \((100)\) and \((100) + (010)\) are standard form expressions, while \((010)\) is not. I expect that the restriction of \(\leq\) to \(OT\) forms a well-ordering. Faery「The weird comparison perhaps works by restricting additions in standard form expressions!! I WIN!!」 = Fundamental Sequence = For an \(s \in T\), I denote by \(L(s)\) the length of \(s\) as formal strings. I define a recursive map \begin{eqnarray*} [ \ ] \colon OT \times \mathbb{N} & \to & OT \\ (s,n) & \mapsto & sn \end{eqnarray*} in the following recursive way: # If \(s = 0\), then \(sn = 0\). # If \(s \neq 0\), then \(sn = \max \{t \in OT \mid t < s \land L(t) < L(s) + 9n\}\). By the definition, \(s \neq 0\) implies \(sn < s\). Faery「Setting FSs is awfully tiresome, so I throw it away in this way!」 = Large Function = I define a recursive map \begin{eqnarray*} (100)_{\bullet}(\bullet) \colon OT \times \mathbb{N} & \to & \mathbb{N} \\ (s,n) & \mapsto & (100)_s(n) \end{eqnarray*} in the following recursive way: # If \(s = 0\), then \((100)_s(n) = n+1\). # If \(s \neq 0\), then \((100)_s(n) = \sum_{m=0}^{n} (100)_{sm}^m(m)\). If the restriction of \(\leq\) to \(OT\) is a well-ordering, then \((100)_{\bullet}(\bullet)\) is total. I define a recursive map \begin{eqnarray*} (100)_{\bullet} \colon \mathbb{N} & \to & OT \\ n & \mapsto & (100)_n \end{eqnarray*} in the following recursive way: # If \(n = 0\), then \((100)_n = (100)\). # If \(n \neq 0\), then \((100)_n = ((100)_{n-1}00)\). The recursive map \((100)_{\bullet}\) gives a limit of the notation system \((OT,\leq)\). Faery「Finally, I've achieved \(\varphi(1,0,0,0) = \varphi(\varphi(\cdots \varphi(1,0,0) \cdots,0,0),0,0)\), haven't I!?」 I note that her analysis is wrong, because it is not an ordinal notation associated to Veblen hierarchy. I expect that the limit of this notation is an OCF-level if it actually works. = Large Number = I submitted the computable large number \((100)_{(100)_{(100)_{(100)}(100)}}(100)\) to the Japanese googological event. Faery oO(Yeah...! Everyone enjoys my ordinal notation associated to Veblen hierarchy...! Zzz...) = Analysis = I show tables of the expectation of the ordinal type \(o(\alpha) \in \Omega\) of the segment \(\{\beta \in OT \mid \beta < \alpha\}\) for an expression \(\alpha \in OT\) without a proof. Since the original system of fundamental sequences is complicated, I exhibit another equivalent system of fundamental sequence, which I will call "modified fundamental sequences" later. Since the analysis of this notation is very difficult for me, the expectation might include so many obvious mistakes. In order to shorten expressions, I employ the following abbreviations: Up to \(\varepsilon_0\) I describe the ordinal types into iterated Cantor normal forms. In this realm, \(o\) preserves the addition. The map \(c \mapsto (00c)\) plays a role analogous to the map \(\gamma \mapsto \omega^{\gamma}\). Up to \(\zeta_0\) I describe the ordinal types into expressions with epsilon function. From this realm on, the correspondence \(o\) does not preserve the addition. For example, \(o((100)+(100))\) isintended to coincide with \(\zeta_0\), which is much greater than \(o((100))+o((100)) = \varepsilon_0 + \varepsilon_0\). The map \(c \mapsto (01c)\) plays a role analogous to the map \(\gamma \mapsto \varepsilon_{\gamma}\). Up to \(\Gamma_0\) I describe the ordinal types into normal forms for Veblen function. The map \((b,c) \mapsto (0bc)\) plays a role analogous to the map \((\beta,\gamma) \mapsto \varphi_{\beta}(\gamma)\), but \(o\) does not preserve the structure because it does not preserve the addition. For example, \(o((00(100)+1))\) is intended to coincide with \(\varphi_{\omega}(0)\), which is much greater than \(\varphi_{0}(o((100)+1)) = \omega^{\varepsilon_0+1}\). Up to BHO I describe the ordinal types into normal forms for Buchholz's function restericted to expressions consisting of \(0\), \(+\), \(\psi_0\), and \(\psi_1\). Up to \(\psi_0(\Omega_3)\) I describe the ordinal types into normal forms for Buchholz's function restericted to expressions consisting of \(0\), \(+\), \(\psi_0\), \(\psi_1\), and \(\psi_2\). WIP Up to BO I describe the ordinal types into normal forms for Buchholz's function. WIP. Up to \(\psi_{0}(\Omega_{\Omega_{\cdot_{\cdot_{\cdot_{\Omega}}}}})\) I describe the ordinal types into normal forms for extended Buchholz's function. WIP. = See Also = * The original Japanese blog post * A source code of the computation of fundamental sequences by rpakr * An analysis by rpakr Category:Blog posts